Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{5z^2 - 30z - 135}{z^2 - 6z - 27} \div \dfrac{z^2 - 9z}{4z^2 + 24z} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{5z^2 - 30z - 135}{z^2 - 6z - 27} \times \dfrac{4z^2 + 24z}{z^2 - 9z} $ First factor out any common factors. $a = \dfrac{5(z^2 - 6z - 27)}{z^2 - 6z - 27} \times \dfrac{4z(z + 6)}{z(z - 9)} $ Then factor the quadratic expressions. $a = \dfrac {5(z + 3)(z - 9)} {(z + 3)(z - 9)} \times \dfrac {4z(z + 6)} {z(z - 9)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac { 5(z + 3)(z - 9) \times 4z(z + 6)} { (z + 3)(z - 9) \times z(z - 9)} $ $a = \dfrac {20z(z + 3)(z - 9)(z + 6)} {z(z + 3)(z - 9)(z - 9)} $ Notice that $(z + 3)$ and $(z - 9)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {20z\cancel{(z + 3)}(z - 9)(z + 6)} {z\cancel{(z + 3)}(z - 9)(z - 9)} $ We are dividing by $z + 3$ , so $z + 3 \neq 0$ Therefore, $z \neq -3$ $a = \dfrac {20z\cancel{(z + 3)}\cancel{(z - 9)}(z + 6)} {z\cancel{(z + 3)}(z - 9)\cancel{(z - 9)}} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $a = \dfrac {20z(z + 6)} {z(z - 9)} $ $ a = \dfrac{20(z + 6)}{z - 9}; z \neq -3; z \neq 9 $